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Complex Eigenvalues

  1. Dec 13, 2004 #1

    cepheid

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    Hello:

    -was solving for the eigenvalues of a matrix. Obtained:
    [tex] \lambda = 1 \pm 2i [/tex]

    -substituted back into matrix to try and solve for the eigenvectors:

    [tex] \left(\begin{array}{cc}2-2i & -2\\4 & -2-2i\end{array}\right) \left(\begin{array}{cc}x_1 \\ x_2 \end{array}\right) = \mathbf{0} [/tex]

    I'm having trouble solving this type of system. I'm just wondering what the general strategy is for tackling these types of matrices.

    Thanks.
     
  2. jcsd
  3. Dec 13, 2004 #2

    shmoe

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    The same procedures as normal will work, don't let the complex numbers intimidate you! Divide the first row by 2-2i and go from there. To make (-2)/(2-2i) nicer to handle, you can multiply the numerator and denominator by the conjugate, 2+2i.

    Once in reduced form, are you able to pick out the eigenvectors?
     
  4. Dec 14, 2004 #3

    cepheid

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    I think I'm going nuts. The matrix I gave was for the eigenvalue 1+2i btw. Here's what I did:

    divide row 1 by 2-2i. Then:

    [tex] \frac{-2}{2-2i} \times \frac{2+2i}{2+2i} = \frac{-4-4i}{4-4i^2} = \frac{-4-4i}{4-4(-1)} = \frac{-2-2i}{4} [/tex]

    The matrix is now:

    [tex] \left(\begin{array}{cc}1 & (-2-2i)/4 \\4 & -2-2i\end{array}\right) \left(\begin{array}{cc}\xi_1 \\ \xi_2 \end{array}\right) = \mathbf{0} [/tex]

    Both rows clearly represent the same equation, namely:

    [tex] \xi_1 = \frac{1+i}{2}\xi_2 [/tex]

    To me, that means:

    [tex] \mathbf{\xi^{(1)}} = \left(\begin{array}{cc} \frac{1+i}{2} \\ 1 \end{array}\right) [/tex]

    But according to Boyce & diPrima:

    [tex] \mathbf{\xi^{(1)}} = \left(\begin{array}{cc} 1 \\ 1-i \end{array}\right) [/tex]

    :confused:
     
  5. Dec 14, 2004 #4

    cepheid

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    Dammit! Nevermind. Those two answers are equivalent!!! I just had to go through a bunch more rigmarole to get it in their form! :mad: Express the ratio between xi1 and xi2 in the reverse way, and multiply by the complex conjugate yet again! (*wonders if there is some way to arrive at it in their form more directly*)
     
  6. Dec 14, 2004 #5

    Galileo

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    Instead of dividing row 1 by 2-2i, it may be easier to multiply by the right amount.

    Muliply by 1+i first. This gives (2-2i)(1+i)=4 and -2(1+i)=-2-2i.
    Dividing by 4, you can see immediately that the reduced form of the matrix is:

    [tex] \left(\begin{array}{cc}1 &
    -\frac{1+i}{2}\\0 & 0\end{array}\right)[/tex]

    So [itex]\frac{1+i}{2} \choose 1[/itex] is solution. Rewrite like you did to get the other form.

    If you want the 1 on top right away, you want a 1 in the upper right corner.
    Start by subtracting the 1st row from the 2nd. You get 2+2i and -2i.
    Multiply 2nd row by -i, you get: 2-2i and -2.
    Dividing the first row by 2, you get:
    [tex] \left(\begin{array}{cc}1-i &
    -1\\0 & 0\end{array}\right)[/tex]

    So a solution is: [itex]1 \choose 1-i[/itex].

    Row reducing complex is usually tedious.
     
  7. Dec 14, 2004 #6

    dextercioby

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    Frien,there's no need to "go nuts"!! :tongue2: Try to put [itex] x_{1} =1[/itex] in the matrix equation and from the first equation u'll find [itex] x_{2}=1-i [/itex].If u put [itex] x_{2} =1 [/itex] in the matrix equation and solving from the second equation u'll find your solution [itex] x_{1}=\frac{1+i}{2} [/itex].U have 2 vectors.See whether they are linear independent.They're not.So,it's irrelevant which u chose in the final solution;

    Daniel.
     
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